Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int getint() {
  int x = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x *= 10, x += (c - '0'), c = getchar();
  return x;
}
vector<pair<int, int> > v[2];
int n, a[2][10010];
void XOR(int i, int j, int idx) {
  v[idx].push_back(make_pair(i, j));
  a[idx][i] ^= a[idx][j];
}
void init() {
  n = getint();
  for (int j = 0; j < 2; j++)
    for (int i = 0; i < n; i++) a[j][i] = getint();
}
void Gauss(int idx) {
  int pre = 0;
  for (int i = 30; i >= 0; i--)
    for (int j = pre; j < n; j++)
      if (a[idx][j] & (1 << i)) {
        if (j != pre) {
          XOR(j, pre, idx);
          XOR(pre, j, idx);
          XOR(j, pre, idx);
        }
        for (int k = 0; k < n; k++)
          if (k != pre && (a[idx][k] & (1 << i))) XOR(k, pre, idx);
        pre++;
        break;
      }
}
void solve() {
  for (int i = 0; i < 2; i++) Gauss(i);
  for (int i = 0; i < n; i++)
    if (a[0][i]) {
      int bs = (1 << 30);
      while ((a[0][i] & bs) == 0) bs >>= 1;
      for (int j = 0; j <= 30 && j < n; j++)
        if ((a[0][j] ^ a[1][j]) & bs) XOR(j, i, 0);
    } else
      break;
  for (int i = 0; i < n; i++)
    if (a[0][i] != a[1][i]) {
      puts("-1");
      return;
    }
  printf("%d\n", (int)v[0].size() + (int)v[1].size());
  for (int i = 0; i < (int)v[0].size(); i++)
    printf("%d %d\n", v[0][i].first + 1, v[0][i].second + 1);
  for (int i = (int)v[1].size() - 1; i >= 0; i--)
    printf("%d %d\n", v[1][i].first + 1, v[1][i].second + 1);
}
int main() {
  init();
  solve();
}